8.3 Weak Acid and Base Equilibria

1. ADiagram 1
   A student who thinks that the molecules of a weak acid do not ionize at all picks this. The diagram shows only HA molecules, but a weak acid does react with water to a small extent, so its solution contains some H₃O⁺ and A⁻ ions.
2. BDiagram 2
   A student who thinks that the reactant and products are present in equal amounts at equilibrium picks this. At equilibrium the forward and reverse rates are equal, not the amounts; for a weak acid the un-ionized molecules greatly outnumber the ions.
3. CDiagram 3
   A student who thinks every acid ionizes completely in water picks this. That is true of strong acids such as HCl; a weak acid ionizes only to a small extent, so most of the HA remains as molecules.
4. DDiagram 4 Correct
   Only a small percentage of the molecules of a weak acid ionize, so most HA molecules remain intact, and each molecule that does ionize gives one H₃O⁺ ion and one A⁻ ion. The diagram with five HA molecules, one H₃O⁺ ion and one A⁻ ion shows this.

Working No calculation. A weak acid ionizes only to a small extent: most HA stays as molecules, and each molecule that ionizes gives one H₃O⁺ and one A⁻. The diagram with five HA molecules, one H₃O⁺ and one A⁻ fits; the others show no ionization, half ionization or complete ionization.

CED 8.3.A.1 · Read this in Fix

1. A1.00
   A student who thinks every acid ionizes completely picks this, taking [H₃O⁺] = 0.10 M. For a weak acid [H₃O⁺] is much less than the initial acid concentration; here it is 7.9 × 10⁻⁴ M.
2. B3.10 Correct
   K_a = 10^−5.20 = 6.3 × 10⁻⁶. With [H₃O⁺] = [A⁻] = x and x small compared with 0.10 M, x² = K_a × 0.10 = 6.3 × 10⁻⁷, so x = 7.9 × 10⁻⁴ M and pH = 3.10. Less than 1% of the HA ionizes, so the approximation holds.
3. C5.20
   A student who thinks the pH of a weak acid solution equals the acid's pK_a picks this. pK_a is a property of the acid; the pH also depends on the initial concentration, through [H₃O⁺] ≈ √(K_a × 0.10).
4. D6.20
   A student who thinks K_a is the fraction of the acid that ionizes picks this: [H₃O⁺] = 6.3 × 10⁻⁶ × 0.10 = 6.3 × 10⁻⁷ M, pH 6.20. K_a is the ratio [H₃O⁺][A⁻]/[HA], so [H₃O⁺] = √(K_a × 0.10) = 7.9 × 10⁻⁴ M.

Working K_a = 10^−5.20 = 6.3 × 10⁻⁶. HA + H₂O ⇌ H₃O⁺ + A⁻; x²/(0.10 − x) = 6.3 × 10⁻⁶. x ≪ 0.10, so x² = 6.3 × 10⁻⁷ and x = [H₃O⁺] = 7.9 × 10⁻⁴ M (0.79% ionized; the quadratic gives the same pH to two decimal places). pH = −log(7.9 × 10⁻⁴) = 3.10. Equivalently pH = ½(pK_a − log[HA]₀) = ½(5.20 + 1.00) = 3.10.

CED 8.3.A.2 · Read this in Fix

1. A[OH⁻] is lower in the NH₃ solution, because most NH₃ molecules remain un-ionized Correct
   NaOH is a strong base: it dissociates completely, so [OH⁻] = 0.10 M. NH₃ reacts with water to produce OH⁻, but only a small percentage of NH₃ molecules do so, so [OH⁻] in the NH₃ solution is much less than 0.10 M.
2. B[OH⁻] is equal in the two solutions, because each base ionizes completely in water
   A student who thinks a weak base ionizes completely, as a strong base does, picks this. Only a small percentage of NH₃ molecules react with water, so [OH⁻] is much less than the 0.10 M in the NaOH solution.
3. C[OH⁻] is zero in the NH₃ solution, because NH₃ contains no OH group to release
   A student who thinks a base must contain OH to produce OH⁻ picks this. NH₃ accepts a proton from water, NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, so its solution does contain OH⁻, although less than the NaOH solution.
4. D[OH⁻] is lower in the NaOH solution, because NaOH stays mostly as NaOH units
   A student who thinks an ionic compound dissolves as units of its formula picks this. NaOH is ionic and a strong base; it dissociates completely into Na⁺ and OH⁻, giving the higher [OH⁻].

Working No calculation. NaOH dissociates completely: [OH⁻] = 0.10 M. NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ proceeds only to a small extent, so [OH⁻] in the NH₃ solution is much less than 0.10 M but not zero.

CED 8.3.A.3 · Read this in Fix

1. A[(CH₃)₂NH₂⁺][OH⁻]/([(CH₃)₂NH][H₂O])
   A student who includes every species in the equation, water too, picks this. Water is the solvent and, like a pure liquid, is left out of the equilibrium expression.
2. B[(CH₃)₂NH]/([(CH₃)₂NH₂⁺][OH⁻])
   A student who thinks K_b measures how much base stays un-ionized picks this, putting [(CH₃)₂NH] in the numerator. An equilibrium expression has the products over the reactants, so K_b is larger for a base that ionizes more.
3. C[(CH₃)₂NH₂⁺][OH⁻]/[(CH₃)₂NH]₀
   A student who thinks the K_b expression uses the initial concentration of the base picks this. Every concentration in K_b is an equilibrium value; [(CH₃)₂NH] is the concentration of base left un-ionized, which only approximately equals [(CH₃)₂NH]₀ when ionization is small.
4. D[(CH₃)₂NH₂⁺][OH⁻]/[(CH₃)₂NH] Correct
   Dimethylamine accepts a proton from water: (CH₃)₂NH + H₂O ⇌ (CH₃)₂NH₂⁺ + OH⁻. K_b has the equilibrium concentrations of the products over the equilibrium concentration of the base, and water, the solvent, is left out.

Working No calculation. (CH₃)₂NH(aq) + H₂O(l) ⇌ (CH₃)₂NH₂⁺(aq) + OH⁻(aq). K_b = equilibrium concentrations of products over reactants, leaving out the solvent water: K_b = [(CH₃)₂NH₂⁺][OH⁻]/[(CH₃)₂NH].

CED 8.3.A.4 · Read this in Fix

1. AMeasure the pH of HA solutions of several known concentrations at varied temperatures
   A student who thinks a test is more thorough when other conditions vary too picks this. The concentrations do vary, but so does the temperature, which changes K_a, so a change in percent ionization could come from either factor.
2. BMeasure the pH of several samples of different volume taken from one HA solution
   A student who confuses the amount of a solution with its concentration picks this. Samples of different volumes taken from one solution all have the same concentration and the same pH, so this procedure cannot show any effect of concentration.
3. CTitrate HA solutions of several known concentrations with NaOH to equivalence
   A student who thinks a base neutralizes only the ionized part of a weak acid picks this. As OH⁻ removes H₃O⁺, more HA ionizes, so all of the acid is neutralized; the volume at equivalence measures the total amount of acid, not the fraction that was ionized.
4. DMeasure the pH of HA solutions of several known concentrations, all at one temperature Correct
   Percent ionization = [H₃O⁺]/[HA]₀ × 100%. Measuring the pH of solutions of several known initial concentrations at one temperature gives [H₃O⁺] = 10^−pH for each, so the percent ionization can be compared across concentrations with only the concentration changed.

Working No calculation. Percent ionization = [H₃O⁺]/[HA]₀ × 100%, with [H₃O⁺] = 10^−pH. The question's independent variable is [HA]₀, so prepare solutions of several known [HA]₀, keep them all at one temperature, and measure each pH.

CED 8.3.A.5 · Read this in Fix

1. A7.00
   A student who thinks a salt solution is always neutral picks this. A⁻ is a weak base with K_b = K_w/K_a = 5.0 × 10⁻¹⁰; it produces OH⁻ by reacting with water, so the solution is basic.
2. B8.85 Correct
   A⁻ is the conjugate base of a weak acid, so it reacts with water: A⁻ + H₂O ⇌ HA + OH⁻. K_b = K_w/K_a = 5.0 × 10⁻¹⁰, [OH⁻] = √(5.0 × 10⁻¹⁰ × 0.10) = 7.1 × 10⁻⁶ M, pOH = 5.15 and pH = 14.00 − 5.15 = 8.85.
3. C5.15
   A student who reports −log[OH⁻] as the pH picks this. 5.15 is the pOH; at 25°C, pH = 14.00 − 5.15 = 8.85.
4. D2.85
   A student who thinks an ion that comes from an acid is itself acidic picks this, using K_a for A⁻: [H₃O⁺] = √(2.0 × 10⁻⁵ × 0.10). A⁻ has already lost its acidic proton; in water it accepts a proton, so the solution is basic.

Working A⁻ + H₂O ⇌ HA + OH⁻. K_b = K_w/K_a = 1.0 × 10⁻¹⁴/2.0 × 10⁻⁵ = 5.0 × 10⁻¹⁰. [OH⁻] = √(5.0 × 10⁻¹⁰ × 0.10) = 7.1 × 10⁻⁶ M. pOH = 5.15; pH = 14.00 − 5.15 = 8.85.

CED 8.3.A.6 · Read this in Fix

1. AH₂ bubbles form at equal rates, because the two acid concentrations are equal
   A student who treats a strong acid and a weak acid at the same concentration as equally acidic picks this. Equal concentrations of acid do not give equal [H₃O⁺]: HCl ionizes completely, while most HA molecules stay un-ionized.
2. BH₂ bubbles form faster in the HCl beaker, because its [H₃O⁺] is greater Correct
   HCl ionizes completely, so [H₃O⁺] in its solution is 0.10 M. HA is weak: only a small percentage of its molecules ionize, so [H₃O⁺] in its solution is much less than 0.10 M. Magnesium reacts with H₃O⁺ to form H₂, and the higher [H₃O⁺] in the HCl beaker gives the faster reaction.
3. CH₂ bubbles form faster in the HA beaker, because it contains more acid molecules
   A student who thinks un-ionized acid molecules are what make a solution acidic picks this. The HA beaker does contain more acid molecules, but the solution's acidity depends on [H₃O⁺], which is far lower in the HA beaker, so bubbling is slower there.
4. DNo H₂ bubbles form in the HA beaker, because a weak acid does not ionize
   A student who thinks a weak acid does not ionize at all picks this. A small percentage of the HA molecules do ionize, so the HA solution contains some H₃O⁺, and bubbles form there too, more slowly than in the HCl beaker.

Working No calculation. HCl is a strong acid: [H₃O⁺] = 0.10 M. HA is weak: only a small percentage ionizes, so [H₃O⁺] ≪ 0.10 M. Mg reacts with H₃O⁺ to form H₂, and the rate is greater where [H₃O⁺] is greater, so bubbles form faster in the HCl beaker; the HA beaker still bubbles, more slowly.

CED 8.3.A.1 · Read this in Fix

1. A[HA] > [A⁻] > [OH⁻] Correct
   Only a small fraction of HA ionizes: [A⁻] = [H₃O⁺] ≈ √(1.0 × 10⁻⁵ × 0.10) = 1.0 × 10⁻³ M, so [HA] ≈ 0.099 M is much greater than [A⁻]. The solution is acidic, so [OH⁻] = K_w/[H₃O⁺] ≈ 1.0 × 10⁻¹¹ M, far smaller than [A⁻].
2. B[A⁻] > [HA] > [OH⁻]
   A student who thinks every acid ionizes completely picks this, expecting most of the acid to be present as ions. Only about 1% of this weak acid is ionized, so [HA], about 0.099 M, remains far greater than [A⁻], about 1.0 × 10⁻³ M.
3. C[HA] = [A⁻] > [OH⁻]
   A student who thinks the reactant and products have equal concentrations at equilibrium picks this. At equilibrium the forward and reverse rates are equal; here [HA], about 0.099 M, is about 100 times [A⁻], about 1.0 × 10⁻³ M.
4. D[HA] > [A⁻] = [OH⁻]
   A student who thinks a weak acid barely changes the pH of water, so that its ions are about as dilute as the ions in pure water, picks this. Here [A⁻] ≈ 1.0 × 10⁻³ M, while [OH⁻] ≈ 1.0 × 10⁻¹¹ M.

Working [H₃O⁺] = [A⁻] = x; x²/(0.10 − x) = 1.0 × 10⁻⁵, so x ≈ √(1.0 × 10⁻⁶) = 1.0 × 10⁻³ M (1.0% ionized). [HA] ≈ 0.10 − 0.0010 = 0.099 M. [OH⁻] = K_w/[H₃O⁺] = 1.0 × 10⁻¹⁴/1.0 × 10⁻³ = 1.0 × 10⁻¹¹ M. So [HA] > [A⁻] > [OH⁻].

CED 8.3.A.1 · Read this in Fix

1. AK_a: HY > HX; pH: HY > HX
   A student who thinks a higher pH means more H₃O⁺ picks this. The HY solution does have more H₃O⁺ ions, but pH = −log[H₃O⁺], so the greater [H₃O⁺] gives the lower pH.
2. BK_a: HY < HX; pH: HY < HX
   A student who thinks a larger K_a means more of the acid stays un-ionized picks this, giving HX the larger K_a. K_a = [H₃O⁺][A⁻]/[HA] is larger for the acid that ionizes to the greater extent, which is HY.
3. CK_a: HY > HX; pH: HY < HX Correct
   Both solutions started with nine acid molecules in the volume shown. In the HY solution two of them have ionized, giving two H₃O⁺ and two Y⁻, while in the HX solution only one has. HY ionizes to the greater extent, so it has the larger K_a; its solution has the greater [H₃O⁺] and therefore the lower pH.
4. DK_a: HY < HX; pH: HY > HX
   A student who thinks un-ionized acid molecules make a solution acidic picks this, because the HX solution has more acid molecules. Acidity depends on [H₃O⁺], which is greater in the HY solution, and the larger K_a belongs to the acid that ionizes more, HY.

Working Each box started with 9 acid molecules. HX box: 8 HX, 1 H₃O⁺, 1 X⁻ (1 of 9 ionized). HY box: 7 HY, 2 H₃O⁺, 2 Y⁻ (2 of 9 ionized). Counting particles as relative concentrations: K_a(HX) ∝ (1)(1)/8 = 0.13; K_a(HY) ∝ (2)(2)/7 = 0.57, so K_a(HY) > K_a(HX). [H₃O⁺] is greater in the HY box, so pH(HY) < pH(HX).

CED 8.3.A.2 · Read this in Fix

1. AIt is smaller, because the [H₃O⁺] calculated from the reading is too low Correct
   A reading 0.20 unit too high gives [H₃O⁺] = 10^{−(pH + 0.20)}, which is 10^−0.20 ≈ 0.63 times the true value. K_a = [H₃O⁺][A⁻]/[HA] ≈ [H₃O⁺]²/(0.100 − [H₃O⁺]), so the calculated K_a is about 0.40 times the true K_a.
2. BIt is larger, because the [H₃O⁺] obtained from the reading is too high
   A student who thinks a higher pH means a higher [H₃O⁺] picks this. pH = −log[H₃O⁺], so a reading that is too high corresponds to an [H₃O⁺] that is too low, and the calculated K_a is too small.
3. CIt is larger, because the [H₃O⁺] calculated from the reading is too low
   A student who thinks a smaller extent of ionization means a larger K_a picks this. K_a has [H₃O⁺] and [A⁻] in the numerator, so an [H₃O⁺] that is too low makes the calculated K_a too small, not too large.
4. DIt is the same, because K_a for HA depends only on the temperature
   A student who thinks a calculated equilibrium constant cannot be affected by a measurement error, because the true K_a depends only on temperature, picks this. The true K_a is fixed at 25°C, but the student's value is calculated from the faulty reading and is about 0.40 times the true value.

Working K_a ≈ [H₃O⁺]²/(0.100 − [H₃O⁺]). A reading 0.20 unit too high gives [H₃O⁺]calc = 10^{−(pH + 0.20)} = 10^−0.20 × [H₃O⁺]true = 0.63 × [H₃O⁺]true. With [H₃O⁺] ≪ 0.100 M, K_a,calc ≈ (0.63)² × K_a,true = 0.40 × K_a,true: smaller.

CED 8.3.A.2 · Read this in Fix

1. A4.0 × 10⁻³
   A student who puts the initial concentration of HX in the K_a expression picks this: (0.020)(0.020)/0.10. Ten HX molecules were present before ionization, but K_a uses the equilibrium concentration, which the diagram shows as eight molecules, 0.080 M.
2. B2.0 × 10⁻²
   A student who thinks K_a equals the equilibrium [H₃O⁺] picks this. K_a is the ratio [H₃O⁺][X⁻]/[HX] = (0.020)(0.020)/0.080.
3. C2.0 × 10⁻¹
   A student who thinks K_a is the fraction of the acid that ionizes picks this: 2 of the 10 HX molecules, 0.20. That fraction is the extent of ionization (20%); K_a is the ratio (0.020)(0.020)/0.080 = 5.0 × 10⁻³.
4. D5.0 × 10⁻³ Correct
   From the diagram, at equilibrium [HX] = 8 × 0.010 M = 0.080 M and [H₃O⁺] = [X⁻] = 2 × 0.010 M = 0.020 M. K_a = (0.020)(0.020)/0.080 = 5.0 × 10⁻³.

Working From the diagram at equilibrium: [HX] = 8 × 0.010 M = 0.080 M; [H₃O⁺] = [X⁻] = 2 × 0.010 M = 0.020 M. K_a = [H₃O⁺][X⁻]/[HX] = (0.020)(0.020)/0.080 = 5.0 × 10⁻³.

CED 8.3.A.2 · Read this in Fix

1. AIt is consistent, because NH₃ ionizes completely to give NH₄⁺ and OH⁻ ions
   A student who thinks every base ionizes completely picks this. The ions shown, NH₄⁺ and OH⁻ in equal numbers, are the right ones, but only a small percentage of NH₃ molecules ionize, so most of the particles should be NH₃ molecules.
2. BIt is not consistent, because NH₃ does not react with water, so no ions should appear
   A student who thinks a weak base does not react with water at all picks this. NH₃ does react with water to a small extent, so a correct diagram shows a few NH₄⁺ and OH⁻ ions; the error in this diagram is that it shows no NH₃ molecules.
3. CIt is not consistent, because the majority of NH₃ molecules should stay un-ionized Correct
   The diagram shows four NH₄⁺ ions and four OH⁻ ions and no NH₃ molecules, as if every NH₃ molecule had ionized. NH₃ is a weak base: only a small percentage of its molecules accept a proton from water, so a model of NH₃(aq) should show mostly NH₃ molecules with a few NH₄⁺ and OH⁻ ions in equal numbers.
4. DIt is not consistent, because NH₃ should form NH₂⁻ and H₃O⁺ ions in water
   A student who thinks any compound containing hydrogen acts as an acid picks this. NH₃ accepts a proton from water, so NH₄⁺ and OH⁻ are the right ions; the error in the diagram is that it shows no un-ionized NH₃ molecules.

Working No calculation. Read the diagram: four N + 4 H particles with a positive charge (NH₄⁺), four O + H particles with a negative charge (OH⁻), and no N + 3 H particles (NH₃). NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ proceeds only to a small extent, so a consistent model shows mostly NH₃ with a few NH₄⁺ and OH⁻ in equal numbers. The diagram shows complete ionization, so it is not consistent.

CED 8.3.A.3 · Read this in Fix

1. A1.00
   A student who thinks every base ionizes completely picks this, taking [OH⁻] = 0.10 M. For a weak base [OH⁻] is much less than the initial base concentration; here it is 2.0 × 10⁻³ M.
2. B2.70 Correct
   K_b = 10^−4.40 = 4.0 × 10⁻⁵. With [OH⁻] = [HB⁺] = x and x small compared with 0.10 M, x² = K_b × 0.10 = 4.0 × 10⁻⁶, so x = 2.0 × 10⁻³ M and pOH = 2.70. Only 2% of B ionizes, so the approximation holds.
3. C4.40
   A student who thinks the pOH of a weak base solution equals the base's pK_b picks this. pK_b is a property of the base; the pOH also depends on the initial concentration, through [OH⁻] ≈ √(K_b × 0.10).
4. D7.00
   A student who thinks a weak base ionizes so little that its solution is nearly neutral picks this. Even 2% ionization of 0.10 M B gives [OH⁻] = 2.0 × 10⁻³ M, far above the 1.0 × 10⁻⁷ M of pure water.

Working K_b = 10^−4.40 = 4.0 × 10⁻⁵. B + H₂O ⇌ HB⁺ + OH⁻; x²/(0.10 − x) = 4.0 × 10⁻⁵. x ≪ 0.10, so x² = 4.0 × 10⁻⁶ and x = [OH⁻] = 2.0 × 10⁻³ M (2.0% ionized; the quadratic gives the same pOH to two decimal places). pOH = −log(2.0 × 10⁻³) = 2.70.

CED 8.3.A.4 · Read this in Fix

1. AThe B₂ solution, because B₂ has the larger pK_b and so is the stronger base
   A student who thinks a larger pK_b means a stronger base picks this. pK_b = −log K_b, so the larger pK_b of B₂ means a smaller K_b and a weaker base, which produces less OH⁻.
2. BThe B₂ solution, because B₁ produces more OH⁻ and so has the lower pH
   A student who treats −log[OH⁻] as the pH picks this. More OH⁻ gives a lower pOH; at 25°C pH = 14.00 − pOH, so the B₁ solution, with more OH⁻, has the higher pH.
3. CThe B₁ solution, because B₁ has the larger K_b and so produces more OH⁻ Correct
   K_b = 10^−pK_b, so B₁ (pK_b 4.80, K_b = 1.6 × 10⁻⁵) is the stronger base, with 100 times the K_b of B₂. At the same initial concentration it produces more OH⁻ (pOH 2.90 against 3.90), so its solution has the higher pH: 11.10 against 10.10.
4. DNeither, because the bases have equal concentrations and so give equal pH
   A student who thinks equal concentrations of two bases give the same pH picks this. The pH also depends on base strength: B₁ has the larger K_b, so at the same concentration it produces more OH⁻.

Working K_b(B₁) = 10^−4.80 = 1.6 × 10⁻⁵; K_b(B₂) = 10^−6.80 = 1.6 × 10⁻⁷. Same [B]₀ = 0.10 M, so B₁ (larger K_b) gives more OH⁻: [OH⁻] ≈ √(1.6 × 10⁻⁶) = 1.3 × 10⁻³ M, pOH 2.90, pH 11.10 (1.3% ionized); B₂: [OH⁻] ≈ 1.3 × 10⁻⁴ M, pOH 3.90, pH 10.10. The B₁ solution has the higher pH.

CED 8.3.A.4 · Read this in Fix

1. A4.00
   A student who treats the weak acid like a strong acid, with [H₃O⁺] proportional to the acid concentration, picks this: a tenfold dilution would then raise the pH by 1. For a weak acid, dilution increases the fraction ionized, so [H₃O⁺] falls only by a factor of √10.
2. B3.50 Correct
   For a weak acid with small ionization, [H₃O⁺] ≈ √(K_a[HA]₀). Diluting tenfold divides [HA]₀ by 10, so [H₃O⁺] is divided by √10 ≈ 3.2, from 1.0 × 10⁻³ M to 3.2 × 10⁻⁴ M, and the pH rises by 0.50, to 3.50.
3. C2.50
   A student who thinks the greater percent ionization after dilution means a greater [H₃O⁺] picks this. The fraction ionized does rise, from 0.10% to 0.32%, but it is a fraction of a concentration ten times smaller, so [H₃O⁺] falls and the pH rises.
4. D3.00
   A student who thinks that, because K_a does not change on dilution, the pH does not change either picks this. K_a stays the same, but with [HA]₀ ten times smaller the [H₃O⁺] that satisfies K_a is smaller, 3.2 × 10⁻⁴ M.

Working [H₃O⁺] = 1.0 × 10⁻³ M at [HA]₀ = 1.0 M, so K_a = (1.0 × 10⁻³)²/(1.0 − 0.0010) = 1.0 × 10⁻⁶. After dilution [HA]₀ = 0.10 M: [H₃O⁺] = √(1.0 × 10⁻⁶ × 0.10) = 3.2 × 10⁻⁴ M (0.32% ionized), pH = 3.50. Proportional reasoning: [H₃O⁺] ∝ √[HA]₀, so a tenfold dilution divides [H₃O⁺] by √10 and raises pH by ½ log 10 = 0.50.

CED 8.3.A.2 · Read this in Fix

1. AIt decreases on dilution, as the pH rises each time [HA]₀ is decreased
   A student who thinks a more dilute acid ionizes less reads the rising pH as less ionization and picks this. The pH does rise, because [H₃O⁺] falls, but [H₃O⁺] falls by a smaller factor than [HA]₀ does, so the fraction of HA ionized rises from 1.3% to 13%.
2. BIt stays the same on dilution, as K_a of HA has one value in all three solutions
   A student who thinks the percent ionization is fixed by K_a picks this. K_a is the same in every solution, but the percent ionization calculated from the data changes, from 1.3% at 0.100 M to 13% at 0.00100 M.
3. CIt increases on dilution, as [H₃O⁺] falls by a smaller factor than [HA]₀ does Correct
   [H₃O⁺] = 10^−pH: 1.3 × 10⁻³ M, 4.2 × 10⁻⁴ M and 1.3 × 10⁻⁴ M. Each tenfold drop in [HA]₀ lowers [H₃O⁺] by only about a factor of 3, so the percent ionization rises: 1.3%, 4.2% and 13%.
4. DIt increases on dilution, as [H₃O⁺] rises each time the measured pH rises
   A student who thinks a higher pH means a higher [H₃O⁺] picks this. As the pH rises, [H₃O⁺] = 10^−pH falls; the percent ionization rises because [H₃O⁺] falls by a smaller factor than [HA]₀ does.

Working [H₃O⁺] = 10^−pH: 1.3 × 10⁻³ M, 4.2 × 10⁻⁴ M, 1.3 × 10⁻⁴ M. Percent ionization = [H₃O⁺]/[HA]₀ × 100%: 1.3%, 4.2%, 13%. Each tenfold dilution lowers [H₃O⁺] by a factor of only about 3, so the percent ionization increases on dilution.

CED 8.3.A.5 · Read this in Fix

1. A1.0% Correct
   K_a = 10^−4.60 = 2.5 × 10⁻⁵. With x = [H₃O⁺] small compared with 0.25 M, x = √(2.5 × 10⁻⁵ × 0.25) = 2.5 × 10⁻³ M, and the percent ionization is 2.5 × 10⁻³/0.25 × 100% = 1.0%.
2. B50%
   A student who thinks the acid and its ions are present in equal amounts at equilibrium picks this, so that half the acid has ionized. K_a is only 2.5 × 10⁻⁵, so at equilibrium the solution is mostly un-ionized HA; [H₃O⁺] is 2.5 × 10⁻³ M, 1.0% of 0.25 M.
3. C0.010%
   A student who thinks [H₃O⁺] equals K_a picks this: 2.5 × 10⁻⁵/0.25 × 100% = 0.010%. K_a = [H₃O⁺][A⁻]/[HA], so [H₃O⁺] = √(K_a × 0.25) = 2.5 × 10⁻³ M.
4. D0.0025%
   A student who thinks K_a is the fraction of the acid that ionizes picks this: 2.5 × 10⁻⁵ × 100% = 0.0025%. The fraction ionized depends on the concentration too: √(K_a/0.25) = 0.010, or 1.0%.

Working K_a = 10^−4.60 = 2.5 × 10⁻⁵. x²/(0.25 − x) = 2.5 × 10⁻⁵; x ≪ 0.25, so x = √(2.5 × 10⁻⁵ × 0.25) = 2.5 × 10⁻³ M = [H₃O⁺] (the quadratic gives 2.49 × 10⁻³ M). Percent ionization = 2.5 × 10⁻³/0.25 × 100% = 1.0%.

CED 8.3.A.5 · Read this in Fix

1. AIt is incorrect, because A⁻ + H₂O ⇌ HA + OH⁻ is not the reverse of the HA ionization Correct
   The reverse of HA + H₂O ⇌ H₃O⁺ + A⁻ is H₃O⁺ + A⁻ ⇌ HA + H₂O, whose K is 1/K_a. The base reaction, A⁻ + H₂O ⇌ HA + OH⁻, is a different reaction; adding it to the acid ionization gives 2H₂O ⇌ H₃O⁺ + OH⁻, so K_a × K_b = K_w and K_b = K_w/K_a.
2. BIt is correct, because reversing a reaction gives the reciprocal of its K value
   A student who takes the reaction of A⁻ with water to be the acid ionization run backward picks this. Reversing a reaction does invert K, but A⁻ + H₂O ⇌ HA + OH⁻ is not the reverse of HA + H₂O ⇌ H₃O⁺ + A⁻: it forms OH⁻ and uses up H₂O.
3. CIt is incorrect, because K_b equals K_a for a conjugate acid-base pair
   A student who thinks an acid and its conjugate base are equally strong picks this. For a conjugate pair K_a × K_b = K_w, so the larger K_a is, the smaller K_b is.
4. DIt is incorrect, because K_a + K_b equals K_w for a conjugate acid-base pair
   A student who carries the sum pK_a + pK_b = pK_w over to the K values picks this. The sum of the logarithms corresponds to a product of the constants: K_a × K_b = K_w.

Working No calculation. Acid ionization: HA + H₂O ⇌ H₃O⁺ + A⁻ (K_a); its reverse, H₃O⁺ + A⁻ ⇌ HA + H₂O, has K = 1/K_a. The base reaction is A⁻ + H₂O ⇌ HA + OH⁻ (K_b), a different reaction. Adding acid ionization and base reaction gives 2H₂O ⇌ H₃O⁺ + OH⁻, so K_a × K_b = K_w and K_b = K_w/K_a.

CED 8.3.A.6 · Read this in Fix

1. AIt is about 1,000 times as great
   A student who thinks K_a is the fraction of the acid that ionizes takes [H₃O⁺] = K_a × [HA]₀: 4.0 × 10⁻⁴ M for HX and 4.0 × 10⁻⁷ M for HY, a ratio of 1,000. Because [H₃O⁺] = [A⁻], [H₃O⁺]² ≈ K_a[HA]₀, so the ratio is √1000 ≈ 30.
2. BIt is about a third as great
   A student who thinks a larger K_a means more of the acid stays un-ionized, as if [HA] were in the numerator, writes [H₃O⁺]² ≈ [HA]₀/K_a and gets √(1.0/4.0 × 10⁻⁴) ÷ √(0.10/4.0 × 10⁻⁶) ≈ 0.32. K_a = [H₃O⁺][A⁻]/[HA], so the acid with the larger K_a and the larger concentration gives the greater [H₃O⁺].
3. CIt is about 30 times as great Correct
   For small ionization [H₃O⁺] ≈ √(K_a[HA]₀): 2.0 × 10⁻² M for HX and 6.3 × 10⁻⁴ M for HY. K_a × [HA]₀ is 1,000 times as great for HX, so [H₃O⁺] is √1000 ≈ 30 times as great.
4. DIt is exactly 10 times as great
   A student who thinks every acid ionizes completely takes [H₃O⁺] = [HA]₀, 1.0 M and 0.10 M, a ratio of exactly 10. Both acids are weak, so [H₃O⁺] depends on K_a as well, through [H₃O⁺] ≈ √(K_a[HA]₀).

Working [H₃O⁺] ≈ √(K_a[HA]₀). HX: √(4.0 × 10⁻⁴ × 1.0) = 2.0 × 10⁻² M (2.0% ionized); HY: √(4.0 × 10⁻⁶ × 0.10) = 6.3 × 10⁻⁴ M (0.63% ionized). Ratio = √(100 × 10) = √1000 ≈ 32, about 30 times (the quadratic gives 31).

CED 8.3.A.2 · Read this in Fix

1. A3.0 × 10⁻²
   A student who thinks K_b is the fraction of the base that ionizes reads 3.0% from the graph and reports that fraction, 0.030, as K_b. K_b is the equilibrium constant [HB⁺][OH⁻]/[B], not the fraction ionized: (6.0 × 10⁻³)²/0.194 = 1.9 × 10⁻⁴.
2. B6.0 × 10⁻³
   A student who thinks [OH⁻] in a weak base solution equals K_b finds [OH⁻] = 0.030 × 0.20 M = 6.0 × 10⁻³ M correctly and reports it as K_b. [OH⁻] must be put into the expression K_b = [HB⁺][OH⁻]/[B], which gives 1.9 × 10⁻⁴.
3. C1.9 × 10⁻⁴ Correct
   At 0.20 M the graph gives 3.0% ionization, so [OH⁻] = [HB⁺] = 0.030 × 0.20 M = 6.0 × 10⁻³ M and 0.194 M of B remains un-ionized. K_b = (6.0 × 10⁻³)²/0.194 = 1.9 × 10⁻⁴.
4. D5.3 × 10⁻³
   A student who takes the percent ionization as the OH⁻ concentration itself uses [OH⁻] = [HB⁺] = 0.030 M and gets (0.030)²/(0.20 − 0.030) = 5.3 × 10⁻³. The 3.0% is a fraction of the 0.20 M of base, so [OH⁻] = 0.030 × 0.20 M = 6.0 × 10⁻³ M and K_b = 1.9 × 10⁻⁴.

Working From the graph, the 0.20 M solution is 3.0% ionized, so [OH⁻] = [HB⁺] = 0.030 × 0.20 M = 6.0 × 10⁻³ M and [B] = 0.20 M − 0.0060 M = 0.194 M. K_b = [HB⁺][OH⁻]/[B] = (6.0 × 10⁻³)²/0.194 = 1.9 × 10⁻⁴.

CED 8.3.A.5 · Read this in Fix

1. A2.0% Correct
   K_a is unchanged, and for small ionization the fraction ionized is about √(K_a/[HA]₀). Lowering [HA]₀ by a factor of 25 raises the percent ionization by a factor of √25 = 5, from 0.40% to 2.0%; [H₃O⁺] itself falls, from 0.010 M to 2.0 × 10⁻³ M.
2. B10%
   A student who thinks that, because K_a does not change, the equilibrium concentrations do not change either keeps [H₃O⁺] at 0.010 M and divides by the new 0.10 M to get 10%. Dilution lowers every concentration and the equilibrium re-establishes: [H₃O⁺] becomes 2.0 × 10⁻³ M, which is 2.0% of 0.10 M.
3. C0.40%
   A student who thinks the percent ionization is fixed by K_a keeps it at 0.40% at every concentration. K_a fixes the ratio [H₃O⁺][A⁻]/[HA], not the fraction ionized; that fraction is about √(K_a/[HA]₀), so it rises to 2.0% when [HA]₀ falls to 0.10 M.
4. D0.080%
   A student who thinks dilution makes a weak acid ionize less, because a more dilute acid is 'weaker', lowers the percent ionization by the factor of 5 and gets 0.080%. Dilution lowers [H₃O⁺] but raises the fraction of HA that is ionized, to 2.0%.

Working In the 2.5 M solution [H₃O⁺] = 0.0040 × 2.5 M = 0.010 M, so K_a = (0.010)²/(2.5 − 0.010) = 4.0 × 10⁻⁵. Because [H₃O⁺] ≈ √(K_a[HA]₀), the fraction ionized, [H₃O⁺]/[HA]₀ ≈ √(K_a/[HA]₀), is inversely proportional to the square root of [HA]₀. The concentration falls by a factor of 2.5/0.10 = 25, so the percent ionization rises by a factor of √25 = 5: 0.40% × 5 = 2.0%. (Check: √(4.0 × 10⁻⁵ × 0.10) = 2.0 × 10⁻³ M, which is 2.0% of 0.10 M; solving the equilibrium expression exactly gives the same result to two significant figures.)

CED 8.3.A.5 · Read this in Fix